Full-Newton step polynomial-time methods for LO based on locally ...

Proof of Theorem 3We apply the composition rule, which is well known.Lemma 3 Let φ i be (κ i , ν i )-SCB’s on D i , <strong>for</strong> i = 1,2. Then φ 1 + φ 2 is a (κ, ν)-SCB<strong>for</strong> D 1 × D 2 , where κ = max {κ 1 , κ 2 } and ν = ν 1 + ν 2 .Since the linear part in φ µ (x, s) is 0-self-concordant, with ν = 0, it suffices to considerf(x, s) = 2n∑j=1ψ b(√xj s jµwhere s = s(x) = Mx + q. In the sequel we will neglect this relation between s and x.Thus we will prove that f(x, s) is (κ, ν)-self-concordant on the set{(x, s) : x ∈ Rn+ , s ∈ R n }+ .),This will imply that f(x, s) is a (κ, ν)-self-concordant barrier function <strong>for</strong> the domain of(SP), which is the intersection of this set and affine space determined by s = Mx + q.We do this by considering each of the terms in the definition separately and then apply thecomposition rules of Lemma 3.23